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Integral of d{x}:
1/(x^3-1)
How do you in partial fractions?:
1/(x^3-1)
Identical expressions
one /(x^ three - one)
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Similar expressions
1/(x^3+1)

The derivative of the function/ 1/(x^3-1)

Derivative of 1/(x^3-1)

The solution

You have entered [src]

  1   
------
 3    
x  - 1

$$\frac{1}{x^{3} - 1}$$

1/(x^3 - 1)

Detail solution

Let $u = x^{3} - 1$ .
Apply the power rule: $\frac{1}{u}$ goes to $- \frac{1}{u^{2}}$
Then, apply the chain rule. Multiply by $\frac{d}{d x} \left(x^{3} - 1\right)$ :
1. Differentiate $x^{3} - 1$ term by term:
  1. Apply the power rule: $x^{3}$ goes to $3 x^{2}$
  2. The derivative of the constant $-1$ is zero.
  The result is: $3 x^{2}$
The result of the chain rule is:

$- \frac{3 x^{2}}{\left(x^{3} - 1\right)^{2}}$
Now simplify:

$- \frac{3 x^{2}}{\left(x^{3} - 1\right)^{2}}$

The answer is:

$- \frac{3 x^{2}}{\left(x^{3} - 1\right)^{2}}$

The graph

Plot the graph f(x)

Plot the graph f'(x)

The first derivative [src]

      2  
  -3*x   
---------
        2
/ 3    \ 
\x  - 1/

$$- \frac{3 x^{2}}{\left(x^{3} - 1\right)^{2}}$$

Simplify

The second derivative [src]

    /          3 \
    |       3*x  |
6*x*|-1 + -------|
    |           3|
    \     -1 + x /
------------------
             2    
    /      3\     
    \-1 + x /

$$\frac{6 x \left(\frac{3 x^{3}}{x^{3} - 1} - 1\right)}{\left(x^{3} - 1\right)^{2}}$$

Simplify

The third derivative [src]

  /           6           3 \
  |       27*x        18*x  |
6*|-1 - ---------- + -------|
  |              2         3|
  |     /      3\    -1 + x |
  \     \-1 + x /           /
-----------------------------
                   2         
          /      3\          
          \-1 + x /

$$\frac{6 \left(- \frac{27 x^{6}}{\left(x^{3} - 1\right)^{2}} + \frac{18 x^{3}}{x^{3} - 1} - 1\right)}{\left(x^{3} - 1\right)^{2}}$$

Simplify

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Derivative of 1/(x^3-1)

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Piecewise:

The solution